The region satisfying the inequalities y - x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Identify the boundary lines.Line $1$: $y - x = 2$ (passes through $(0, 2)$ and $(-2, 0)$).Line $2$: $x + y = 5$ (passes through $(0, 5)$

Vedclass · Accepted Answer

Bounded

Vedclass · Answer

(B) Step $1$: Identify the boundary lines.Line $1$: $y - x = 2$ (passes through $(0, 2)$ and $(-2, 0)$).Line $2$: $x + y = 5$ (passes through $(0, 5)$ and $(5, 0)$).Line $3$: $x = 0$ ($y$-axis).Line $4$: $y = 0$ ($x$-axis).Step $2$: Determine the feasible region.The region $y - x \geq 2$ lies above the line $y = x + 2$.The region $x + y \leq 5$ lies below the line $x + y = 5$.The region $x \geq 0$ and $y \geq 0$ is the first quadrant.Step $3$: Find the intersection points.Intersection of $y - x = 2$ and $x + y = 5$: Adding the equations gives $2y = 7$, so $y = 3.5$. Then $x = 1.5$. Point is $(1.5, 3.5)$.Intersection of $y - x = 2$ and $x = 0$ is $(0, 2)$.Intersection of $x + y = 5$ and $x = 0$ is $(0, 5)$.Step $4$: The region is a triangle with vertices $(0, 2)$, $(0, 5)$, and $(1.5, 3.5)$. Since the region is enclosed by these lines, it is bounded.

The region satisfying the inequalities $y - x \geq 2$, $x + y \leq 5$, $x \geq 0$ and $y \geq 0$ is

Similar Questions

Solve the following system of inequalities graphically: $x < 1, y < 0, x \geq -3, x + y \geq 0$.

Solve the following system of inequalities graphically: $x > 0, y > 0, x \leq 3, y \leq 2$.

The set $\{x \in R: \frac{14x}{x+1} - \frac{9x-30}{x-4} < 0\}$ is equal to

Solve the following system of inequalities graphically: $x+y < 2, x > 0, y > 1$.

The shaded region in the figure is the solution set of the inequations:

Vedclass Test Series

Exam Paper Generator

Online Exam Module